The equation of the curve through the point (1,0) which satisfies the differential equatoin `(1+y^(2))dx-xydy=0`, is

Options

A. `x^(2)+y^(2)=4`
B. `x^(2)-y^(2)-=`
C. `2x^(2)+y^(2)=2`
D. none of these

Answer

To solve the differential equation \((1+y^2)dx - xy dy = 0\) and find the equation of the curve that passes through the point \((1,0)\), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given differential equation:
\[ (1+y^2)dx - xy dy = 0 \]
Rearranging gives:
\[ (1+y^2)dx = xy dy \]

### Step 2: Find \(\frac{dy}{dx}\)
Dividing both sides by \(dx\) and \(xy\) gives:
\[ \frac{dy}{dx} = \frac{1+y^2}{xy} \]

### Step 3: Separate Variables
We can separate the variables \(x\) and \(y\):
\[ \frac{y \, dy}{1+y^2} = \frac{1}{x} \, dx \]

### Step 4: Integrate Both Sides
Now we integrate both sides:
\[ \int \frac{y \, dy}{1+y^2} = \int \frac{1}{x} \, dx \]

#### Left Side Integration
For the left side, we can use the substitution \(t = 1 + y^2\), then \(dt = 2y \, dy\) or \(\frac{dt}{2} = y \, dy\):
\[ \int \frac{y \, dy}{1+y^2} = \frac{1}{2} \int \frac{dt}{t} = \frac{1}{2} \ln |t| + C_1 = \frac{1}{2} \ln(1+y^2) + C_1 \]

#### Right Side Integration
The right side integrates to:
\[ \int \frac{1}{x} \, dx = \ln |x| + C_2 \]

### Step 5: Combine the Results
Equating both integrals gives:
\[ \frac{1}{2} \ln(1+y^2) = \ln |x| + C \]
where \(C = C_2 - C_1\).

### Step 6: Simplify the Equation
Multiplying through by 2:
\[ \ln(1+y^2) = 2 \ln |x| + 2C \]
Using the property of logarithms, we can rewrite this as:
\[ \ln(1+y^2) = \ln(x^2) + \ln(e^{2C}) = \ln(kx^2) \]
where \(k = e^{2C}\).

### Step 7: Exponentiate Both Sides
Exponentiating both sides gives:
\[ 1+y^2 = kx^2 \]

### Step 8: Rearranging to Find the Curve Equation
Rearranging gives:
\[ kx^2 - y^2 = 1 \]

### Step 9: Use the Initial Condition
We know the curve passes through the point \((1,0)\):
\[ k(1)^2 - (0)^2 = 1 \implies k = 1 \]
Thus, the equation of the curve is:
\[ x^2 - y^2 = 1 \]

### Final Answer
The equation of the curve is:
\[ x^2 - y^2 = 1 \]